find $\int _\gamma \frac{1}{z+\frac {1}{2}}dz$

Question

I'm asked to find $$\int _\gamma \frac{1}{z+\frac {1}{2}}$$ where $\gamma (t)=e^{it}, 0\leq t\leq 2\pi$.

To do this I deal with two different logarithms, one without the negative imaginary axis which I call $\log _1$ and another without the positive imaginary axis, $\log _2$.

Then $$\int _\gamma \frac{1}{z+\frac {1}{2}}=\log _1(z+\frac {1}{2})\bigg|^{z=-1}_{z=1}+\log _2(z+\frac {1}{2})\bigg|^{z=1}_{z=-1}=\\ (\log _1(-\frac {1}{2})-\log _1(\frac{3}{2}))+(\log _2(\frac {3}{2})-\log _2(-\frac{1}{2}))$$

now

• $\log _1(-\frac {1}{2})=\log (\frac{1}{2})+i\pi$
• $\log _1(\frac {3}{2})=\log (\frac{3}{2})+0$
• $\log _2(\frac {3}{2})=\log (\frac{3}{2})+\color{green}{2i\pi}$, maybe $0$ instead of green??
• $\log _2(-\frac {1}{2})=\log (\frac{1}{2})+i\pi $

so if green is right the integral is $2\pi i$ if instead of green it is $0$ the integral is $0$.

I think green is right.

can you please tell me the right answer?

Answer 1

Your $\;\log_2(3/2)\;$ is wrong, imo:

$$\log_1\left(-\frac12\right)-\log_1\frac32+\log_2\left(\frac32\right)-\log_2\left(-\frac12\right)=$$

$$=\log\frac12+\pi i-\log\frac32+\log\frac32+2\pi i-\log\frac12-\pi i=2\pi i$$

Answer 2

Let $f(z)=1$. Using Cauchy's Integral Formula, we know that

$$ \int_\gamma\frac{f(z)}{z+\frac{1}{2}}dz=2\pi i\cdot f\left(-\frac{1}{2}\right)=2\pi i $$